Non-Euclidean CRFs

Discussion of known convex regular-faced polytopes, including the Johnson solids in 3D, and higher dimensions; and the discovery of new ones.

Non-Euclidean CRFs

Postby mr_e_man » Sat Dec 01, 2018 11:56 pm

Are the Johnson solids the same in spherical or hyperbolic space? I would expect that the curvature, changing the angles, adds or removes some possibilities. Of course, the more symmetric solids (like the cube) exist in all three spaces, though the shape depends on size.

In hyperbolic space, a horosphere is both convex and intrinsically flat. So Euclidean plane tilings with regular tiles can be made convex. I suppose these should be excluded from CRFs, simply because they're infinite. Not only does each tiling have infinitely many tiles, but there are (uncountably!*) infinitely many different types of tilings.

Hyperbolic n-space can be embedded in (n,1) psEuc space (as the hyperboloid model); then a horosphere becomes a circular paraboloid. And the square tiling becomes a paraboloidal polyhedron: take z = w - 1 = (x2 + y2)/2 with x and y integers, to generate the vertices, then take the convex hull. So these Euclidean tilings can be made convex in flat space, though the angles and symmetries are non-Euclidean. In particular, a translation becomes a parabolic rotation.

We could also consider tilings with vertices on the other unit hyperboloid (the single-sheet one), but those would not be convex.

I'm also interested in more general psEuc spaces like (2,2). This is a 4-dimensional space with a null cone made of rays from the origin through a Clifford torus: x2 + y2 - z2 - w2 = 0. I wonder what types of symmetries a polytope could have here. The orthogonal group O(2,2) contains the general linear group GL(2) which is represented by all invertible 2x2 matrices.

*A Euclidean tiling can be made by stacking rows on each other, where each row is a strip of triangles or squares with the same edge lengths. A strip of squares represents a 0, and a strip of triangles represents a 1; the whole tiling represents a number in binary, with infinitely many digits. There are uncountably many such numbers, thus uncountably many such tilings. See https://en.wikipedia.org/wiki/Cantor%27 ... l_argument .
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Re: Non-Euclidean CRFs

Postby Marek14 » Wed Dec 05, 2018 8:16 am

In spherical space, Johnson solids are the same, but hyperbolic space has infinitely many.

One way to see it is to consider stacking cubes in a tower. In Euclidean space, these are not Johnson solids because double of the dihedral angle of cube is 180 degrees. But in hyperbolic space, any cube will have smaller dihedral angle and so a tower of cube of any height will still be convex.

Another example are pyramids and bipyramids. In Euclidean space, you can only have pentagonal pyramid as a Johnson solid, but in hyperbolic space, you are not limited -- any pyramid can be made into a Johnson solid.

Finally, you can make "lenses": take a portion of a horosphere or pseudosphere, tile it in any way you want, and then just glue two such portions with matching vertices together.
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Re: Non-Euclidean CRFs

Postby mr_e_man » Thu Dec 06, 2018 9:48 am

:nod: The tower of cubes can also be thought of as a tiling (or truncation) of a cylinder.

...So the number of finite hyperbolic CRFs is at least countably infinite. I believe it must be countable: For any number n, we can make a finite list of all possible polyhedra with at most n faces, each face having at most n edges; then augment the list with that for n+1, etc. This list will eventually reach any such polyhedron, so they are countable.

And I've already shown, by "Euclidean" tilings of a horosphere, that the number of infinite hyperbolic CRFs is uncountably infinite.

The 3D Euclidean regular-celled honeycombs, which have only 3 types of cells (cube, tet, oct), are also uncountable. The argument is similar to the 2D case. Instead of rows of triangles and squares, we use the planes between layers of the octet honeycomb; a plane where octs connect to tets is a 0, and a plane where octs connect to octs is a 1.

Finally, you can make "lenses": take a portion of a horosphere or pseudosphere, tile it in any way you want, and then just glue two such portions with matching vertices together.


Could you show a picture of this, or further explanation? I think it would be rare to be able to match the vertices.
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Re: Non-Euclidean CRFs

Postby Marek14 » Thu Dec 06, 2018 10:01 am

mr_e_man wrote::nod: The tower of cubes can also be thought of as a tiling (or truncation) of a cylinder.

...So the number of finite hyperbolic CRFs is at least countably infinite. I believe it must be countable: For any number n, we can make a finite list of all possible polyhedra with at most n faces, each face having at most n edges; then augment the list with that for n+1, etc. This list will eventually reach any such polyhedron, so they are countable.

And I've already shown, by "Euclidean" tilings of a horosphere, that the number of infinite hyperbolic CRFs is uncountably infinite.

The 3D Euclidean regular-celled honeycombs, which have only 3 types of cells (cube, tet, oct), are also uncountable. The argument is similar to the 2D case. Instead of rows of triangles and squares, we use the planes between layers of the octet honeycomb; a plane where octs connect to tets is a 0, and a plane where octs connect to octs is a 1.

Finally, you can make "lenses": take a portion of a horosphere or pseudosphere, tile it in any way you want, and then just glue two such portions with matching vertices together.


Could you show a picture of this, or further explanation? I think it would be rare to be able to match the vertices.


Well, bipyramids and cupolas are example (hexagonal cupola would have its triangles/squares/hexagons inscribed into horosphere, higher cupolas into pseudosphere).

And I thought of this: If you take a piece of convex triangular pseudohedron, doesn't it have enough degrees of freedom to press the edges along a plane and then reflect it and join with a second copy?
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Re: Non-Euclidean CRFs

Postby wendy » Fri Dec 07, 2018 8:33 am

To get an idea of the infiniteness of hyperbolic CRF, consider something like {5,3,5}. If you join two cells together, the resulting polytope has 11 pentagons, and is convex.

As long as you don't have three dodecahedra at an edge, you can keep adding dodecahedra to your heart's content, and they would be convex-regular-face polytopes. Every imaginable way of stacking these in this way, gives rise to a convex polyhedron.
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Re: Non-Euclidean CRFs

Postby mr_e_man » Fri Dec 07, 2018 10:09 pm

Marek14 wrote: And I thought of this: If you take a piece of convex triangular pseudohedron, doesn't it have enough degrees of freedom to press the edges along a plane and then reflect it and join with a second copy?


Maybe, but it's not clear to me.

wendy wrote: To get an idea of the infiniteness of hyperbolic CRF...


And yet it's the smallest possible infinity! :D

Anyway, thanks for the examples: the dodecs and pyramids and such.

mr_e_man wrote: I'm also interested in more general psEuc spaces like (2,2). This is a 4-dimensional space with a null cone made of rays from the origin through a Clifford torus: x2 + y2 - z2 - w2 = 0. I wonder what types of symmetries a polytope could have here. The orthogonal group O(2,2) contains the general linear group GL(2) which is represented by all invertible 2x2 matrices.


Is anyone familiar with these "indefinite orthogonal groups"? A discrete subgroup could take one point and generate the vertices of a polytope; or take one CRF and possibly generate a higher-dimensional CRF, if the generated edges and faces align correctly.
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Re: Non-Euclidean CRFs

Postby mr_e_man » Sat Dec 08, 2018 9:09 pm

A simple example is the group generated by two orthogonal hyperbolic rotations, just like a duoprism is generated by two orthogonal circular rotations. This would produce vertices at (x,y,z,w) = (A sinh mα, B sinh nβ, A cosh mα, B cosh nβ), where m and n vary through the integers. Using the psEuc metric (which gives the dot product some minus signs), the distance between the adjacent vertices (m,n) and (m+1,n) is 2A sinh α/2. If we want all edge lengths to be equal, then this requires 2A sinh α/2 = 2B sinh β/2.

For simplicity, let's take A=B=sinh α=sinh β=1. It follows that cosh α=sqrt2, which I'll call r to make the coordinates more legible. The hyperbolic functions decompose by sinh(a+b) = sinh a cosh b + cosh a sinh b, and cosh(a+b) = cosh a cosh b + sinh a sinh b, so all the vertices (m,n) are

(0,0): (0, 0, 1, 1). (0,±1): (0, ±1, 1, r). (0,±2): (0, ±2r, 1, 3). (0,±3): (0, ±7, 1, 5r). (0,±4): (0, ±12r, 1, 17). (0,±5): (0, ±41, 1, 29r). ...
(1,0): (1, 0, r, 1). (1,±1): (1, ±1, r, r). (1,±2): (1, ±2r, r, 3). (1,±3): (1, ±7, r, 5r). (1,±4): (1, ±12r, r, 17). (1,±5): (1, ±41, r, 29r). ...
(2,0): (2r, 0, 3, 1). (2,±1): (2r, ±1, 3, r). (2,±2): (2r, ±2r, 3, 3). (2,±3): (2r, ±7, 3, 5r). (2,±4): (2r, ±12r, 3, 17). (2,±5): (2r, ±41, 3, 29r). ...
...

In general, if vertex (m,n) has coordinates (x,y,z,w), then vertex (m+1,n) has coordinates (xr+z, y, x+zr, w), and vertex (m,n+1) has coordinates (x, yr+w, z, y+wr).

Of course, there are edges and square faces between these adjacent vertices. The other faces are "horizontal" apeirogons with varying m and fixed n, and "vertical" apeirogons with fixed m and varying n. The cells are apeirogonal prisms.
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